Gọi x,y lần lượt là số mol của Al, Fe

n_H2 = \(\dfrac{8,96}{22,4}\)=0,4 mol

Pt: 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

......x.................................0,5x...........1,5x

.....Fe + H₂SO₄ --> FeSO₄ + H₂

.......y..........................y............y

Ta có hệ pt:

 {27x+56y=11

   1,5x+y=0,4

⇔x=0,2, y=0,1

% m_Al = \(\dfrac{0,2.27}{11}\).100%=49,1%

% m_Fe = \(\dfrac{0,1.56}{11}\).100%=50,9%

m_Al2(SO4)3 = 0,5x . 342 = 0,5 . 0,2 . 342 = 34,2 (g)

m_FeSO4 = 152y = 152 . 0,1 = 15,2 (g)

Gọi CTTQ: M_xO_y

Pt: M_xO_y + yH₂ --to--> xM + yH₂O

\(\dfrac{0,4}{y}\)<-------0,4

Ta có: 232,2=\(\dfrac{0,4}{y}\)(56x+16y)

⇔23,2=\(\dfrac{22,4x}{y}\)+6,4

⇔\(\dfrac{22,4x}{y}\)=16,8

⇔22,4x=16,8y

⇔x:y=3:4

Vậy CTHH của oxit: Fe₃O₄

\(n_{H_2\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{1,792}{22,4}=0,08\left(mol\right)\)

đặt \(\left\{{}\begin{matrix}n_{Zn}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\)

\(PTHH:Zn+H_2SO_4->ZnSO_4+H_2\)

tỉ lệ           1    :     1        :       1          : 1

n(mol)       a---->a------------>a---------->a (1)

\(PTHH:2Al+3H_2SO_4->Al_2\left(SO_4\right)_3+3H_2\)

tỉ lệ            2  :         3        ;           1           :   3

n(mol)      b-------->3/2b----->1/2b------------>3/2b (2)

Từ (1) và (2) ta có

\(\left\{{}\begin{matrix}65a+27b=3,79\\a+\dfrac{3}{2}b=0,08\end{matrix}\right.=>\left\{{}\begin{matrix}a=0,05\left(mol\right)\\b=0,02\left(mol\right)\end{matrix}\right.\)

\(=>\left\{{}\begin{matrix}m_{Zn}=n\cdot M=0,05\cdot65=3,25\left(g\right)\\m_{Al}=n\cdot M=0,02\cdot27=0,54\left(g\right)\end{matrix}\right.\)

\(=>\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{3,25\cdot100\%}{3,79}\approx85,75\%\\\%m_{Al}=100\%-85,75\approx14,25\%\end{matrix}\right.\)

với (1) thì

\(n_{H_2SO_4\left(1\right)}=a=0,05\left(mol\right)\)

với (2) thì

\(n_{H_2SO_4\left(2\right)}=\dfrac{3}{2}b=\dfrac{3}{2}\cdot0,02=0,03\left(mol\right)\)

\(=>m_{H_2SO_4}=\left(0,05+0,03\right)\cdot98=7,84\left(g\right)\)

PTHH: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)

            \(Cu+2H_2SO_{4\left(đ\right)}\xrightarrow[]{t^o}CuSO_4+SO_2\uparrow+2H_2O\)

Ta có: \(n_{SO_2}=\dfrac{0,112}{22,4}=0,005\left(mol\right)=n_{Cu}\)

\(\Rightarrow m_{Cu}=0,005\cdot64=0,32\left(g\right)\) \(\Rightarrow m_{Mg}=10,6-0,32=10,28\left(g\right)\)

Giả sử: \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\)

⇒ 24x + 27y = 7,8 (1)

Ta có: \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

BT e, có: 2x + 3y = 0,8 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,2\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Mg}=0,1.24=2,4\left(g\right)\\m_{Al}=0,2.27=5,4\left(g\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{2,4}{7,8}.100\%\approx30,77\%\\\%m_{Al}\approx69,23\%\end{matrix}\right.\)

b, BTNT Mg và Al, có:

n_MgCl2 = n_Mg = 0,1 (mol)

 n_AlCl3 = n_Al = 0,2 (mol)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{MgCl_2}=\dfrac{0,1.95}{0,1.95+0,2.133,5}.100\%\approx26,24\%\\\%m_{AlCl_3}\approx73,76\%\end{matrix}\right.\)

Bạn tham khảo nhé!

^{trả lời nhanh cho em ạ}

PTHH: 

Cu + H₂SO₄ ---x--->

Fe + H₂SO₄ ---> FeSO₄ + H₂ (1)

2Cu + O₂ ---t^o---> 2CuO (2)

Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Theo PT₍₁₎: \(n_{Fe}=n_{H_2}=0,3\left(mol\right)\)

=> \(m_{Fe}=0,3.56=16,8\left(g\right)\)

Ta có: \(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)

Theo PT₍₂₎: \(n_{Cu}=n_{CuO}=0,05\left(mol\right)\)

=> \(m_{Cu}=0,05.64=3,2\left(g\right)\)

=> \(\%_{m_{Cu}}=\dfrac{3,2}{3,2+16,8}.100\%=16\%\)

\(\%_{m_{Fe}}=100\%-16\%=84\%\)

nH2= 0,12 mol

Fe + 2HCl = FeCl2 + H2 

=> nFe = nH2 = 0,12 mol

=> mFe = 0,12*56 = 6,72 gam

=%Fe = (6,72*100%):10 = 67,2%
=> % Cu = 100% -67,2% = 32,8%